A charge Q accumulates on the hollow metallic dome,of radius R,of a Van de Graaff generator.A probe measures the electric field strength at various points outside the sphere's surface.By what factor will the electric field value at the 2R distance be changed if the charge value were increased to (5/3)Q? A) 3/5 B) (5/3)² C) 5/3 D) (3/5)²

Question

Quizplus · Accepted Answer

The electric field strength due to a point charge Q at distance r from it is given by:

E = kQ/r^2

where k is Coulomb's constant.

In this case, the charge on the metallic dome is Q and its radius is R. Thus, the electric field just outside the surface of the dome (at a distance R) is:

E1 = kQ/R^2

When we move to a distance of 2R from the center of the dome, the electric field due to the charge Q is:

E2 = kQ/(2R)^2 = kQ/4R^2 = (1/4)E1

Now, if the charge on the dome is increased to (5/3)Q, the electric field at a distance of 2R is:

E2' = k(5/3)Q/(2R)^2 = (5/12)E1

Therefore, the factor by which the electric field value at the 2R distance changes is:

E2'/E2 = (5/12)E1 / (1/4)E1 = (5/12) * 4 = 5/3

Thus, the answer is choice C.

A Charge Q Accumulates on the Hollow Metallic Dome,of Radius